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Question

An αparticle of energy 3 MeV is scattered through 180 by gold nucleus. The distance of the closest approach is of the order of -

The atomic number of gold is 79.

A
1011 cm
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B
1016 cm
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C
1013 cm
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D
1019 cm
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Solution

The correct option is A 1011 cm
From law of conservation of mechanical energy,

KEi+PEi=KEf+PEf

3×106×1.6×1019+0=0+9×109×(79×1.6×1019)×(2×1.6×1019)r

r=7.6×1014 m=7.6×1012 cm1011 cm

Therefore, the distance of the closest approach is of the order of 1011 cm.

Hence, option (A) is the correct answer.

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